The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $20.8$ years; the standard deviation is $4.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $24.9$ years.
Answer: $20.8$ $16.7$ $24.9$ $12.6$ $29$ $8.5$ $33.1$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $20.8$ years. We know the standard deviation is $4.1$ years, so one standard deviation below the mean is $16.7$ years and one standard deviation above the mean is $24.9$ years. Two standard deviations below the mean is $12.6$ years and two standard deviations above the mean is $29$ years. Three standard deviations below the mean is $8.5$ years and three standard deviations above the mean is $33.1$ years. We are interested in the probability of a porcupine living less than $24.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the porcupines will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $16.7$ years and the other half $({16\%})$ will live longer than $24.9$ years. The probability of a particular porcupine living less than $24.9$ years is ${68\%} + {16\%}$, or $84\%$.